Added Solution2!

c5ba8877 · Maximilian Vitz · dfd9db9d · c5ba8877
Commit c5ba8877 authored Oct 28, 2021 by Maximilian Vitz
--- a/Week2/Solution2.ipynb
+++ b/Week2/Solution2.ipynb
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### 1 Exspectation Value and Variance of a Sample"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "1)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "See lecture slide 46."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$\\left< V(x) \\right> = \\left< \\frac{1}{N} \\sum_{i} (x_{i}-\\mu_{x})^{2} \\right> = \\frac{1}{N} \\left<  \\sum_{i} x_{i}^{2} - 2 x_{i} \\mu_{x} + \\mu_{x}^{2} \\right> = \\frac{1}{N}  \\left( \\sum_{i} \\left< x_{i}^{2} \\right> - 2 \\left< x_{i} \\right> \\mu_{x} \\right) + \\mu_{x}^{2} = \\left< x^{2} \\right> - 2 \\mu_{x}^{2} + \\mu_{x}^{2} = \\left< x^{2} \\right> - \\left< x \\right>^{2} = \\sigma_{x}^{2}$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "By fixing a parameter for the measured mean like it was done in 1) one has to apply a correction factor. If the true mean is known this is not neccessary any more."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### 2 Poisson Distribution"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$P_{k} (T) = \\frac{1}{k!} e^{- \\lambda T} (\\lambda T)^{k}$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "(lhs) $\\dot{P}_{k}(T) = \\frac{1}{k!} e^{- \\lambda T} (- \\lambda (\\lambda T)^{k} + k \\lambda^{k} T^{k-1} )$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "(rhs) $\\lambda [P_{k-1}(T) - P_{k}(T)] = \\lambda \\left[ \\frac{e^{-\\lambda T}}{(k-1)!} (\\lambda T)^{k-1} - \\frac{e^{-\\lambda T}}{k!} (\\lambda T)^{k} \\right] = \\frac{1}{k!} e^{- \\lambda T} (- \\lambda (\\lambda T)^{k} + k \\lambda^{k} T^{k-1} )$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$\\Rightarrow \\dot{P}_{k}(T) = \\lambda [P_{k-1}(T) - P_{k}(T)]$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Differential Equation was derived by comparing small time intervall with a Taylor expansion, see lecture slide 62."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### 3 Bionomial- and Poisson Distribution"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "1)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Bionomial Distribtion for probobility for success with N trials!"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$B(k; N, p) = \\left(\\begin{array}{c} \n",
+    "N \\\\ k\n",
+    "\\end{array}\\right) \\cdot p^{k} \\cdot (1-p)^{N-k}$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "N = 60"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "p = 0.01"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "k = 0"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$\\Rightarrow B(0; 60, 0.01) = \\left(\\begin{array}{c} \n",
+    "60 \\\\ 0\n",
+    "\\end{array}\\right) \\cdot 0.01^{0} \\cdot (1-0.01)^{60-0} = 1 \\cdot 1 \\cdot (0.99)^{60} = 54.7 \\%$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Poisson Distribution for event which occur with an average event rate $\\lambda$."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$P(k, \\lambda) = \\frac{\\lambda^{k}}{k!} e^{- \\lambda}$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$\\lambda$ = $0.01 \\cdot 60 = 0.6$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "k = 0"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$P(0,0.6) = \\frac{0.6^{0}}{0!} e^{- 0.6} = 1 \\cdot e^{-0.6} = 54.9 \\%$ "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### 4 Poisson Distribution, Search for free Quarks"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "1)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$\\sum_{i=1}^{110} P(i;229) = \\sum_{i=1}^{110} \\frac{229^{i}}{i!} e^{-229}= 1.6 \\cdot 10^{-18} = p_{se}$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Very low propability, very unlikely to happen!"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "N = 55000"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$p_{me} = N \\cdot p_{se} = 8.9 \\cdot 10^{-14}$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "3)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$\\sum_{i=1}^{28} P(i;57) = \\sum_{i=1}^{28} \\frac{57^{i}}{i!} e^{-57}= 6.7 \\cdot 10^{-6}$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Much more likely to happend!"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "4)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Combine both Poisson Distributions to achieve the new probability:"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$\\mu = 4$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$\\mu \\lambda = 229$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$\\sum_{i=1}^{110} \\sum_{N=0}^{\\infty} P(i;N\\mu) \\cdot P(N;\\lambda) = \\sum_{i=1}^{110} \\sum_{N=0}^{\\infty} \\frac{N \\mu^{i}}{i!} e^{-N \\mu} \\cdot \\frac{\\lambda^{N}}{N!} e^{-\\lambda} = 4.2 \\cdot 10^{-5}$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Even more likely to happen."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### 5 Why are soccer results more random than handball results?"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "1)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$\\lambda_{1} = 1$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$\\lambda_{2} = 2$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$\\sum_{i=1}^{\\infty} \\sum_{l=0}^{k-1} P(k;\\lambda_{1}) \\cdot P(l;\\lambda_{2})= \\sum_{i=1}^{\\infty} \\sum_{l=0}^{k-1}  \\frac{\\lambda_{2}^{k}}{k!} e^{-\\lambda_{1}} \\cdot \\frac{\\lambda_{2}^{l}}{l!} e^{-\\lambda_{2}} = 18.3 \\%$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$\\lambda_{1} = 10$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$\\lambda_{2} = 20$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$\\sum_{i=1}^{\\infty} \\sum_{l=0}^{k-1} P(k;\\lambda_{1}) \\cdot P(l;\\lambda_{2})= 2.6 \\%$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Summing up over all possible combinations, i=1 as at least one goal is needed to win a game."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### 5 Python Script (thanks to Joep Geuskens)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "from scipy.stats import poisson\n",
+    "import numpy as np"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def func(l=1):\n",
+    "    arr  = np.array([poisson.cdf(k-1, 2*l)*poisson.pmf(k,l) for k in range(1,l*10)])\n",
+    "    p    = np.sum(arr)\n",
+    "    diff = arr[-1]/p          # relative size of the last term compared to the total probability\n",
+    "    print(f\"P(B>A)={p:.4f}\")\n",
+    "    print(f\"Diff={diff:.2g}\") # should be sufficiently small"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1)\n",
+      "P(B>A)=0.1826\n",
+      "Diff=5.6e-06\n",
+      "\n",
+      "2)\n",
+      "P(B>A)=0.0258\n",
+      "Diff=1.9e-60\n"
+     ]
+    }
+   ],
+   "source": [
+    "print(\"1)\")\n",
+    "func(1)\n",
+    "print(\"\\n2)\")\n",
+    "func(10)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### 6 Multidimensional Gaussian"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$B = \n",
+    "\\frac{1}{\\sigma_{1}^{2}\\sigma_{2}^{2}- \\rho^{2}\\sigma_{1}^{2}\\sigma_{2}^{2}}\n",
+    "\\left(\\begin{array}{cc} \n",
+    "\\sigma_{2}^{2} & -c\\\\\n",
+    "-c & \\sigma_{1}^{2}\n",
+    "\\end{array}\\right) = \n",
+    "\\frac{1}{\\sigma_{1}^{2}\\sigma_{2}^{2} \\cdot (1- \\rho^{2})}\n",
+    "\\left(\\begin{array}{cc} \n",
+    "\\sigma_{2}^{2} & -c\\\\\n",
+    "-c & \\sigma_{1}^{2}\n",
+    "\\end{array}\\right)$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$c = \\rho x_{1} x_{2}$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$det B = \\frac{1}{\\sigma_{1}^{2}\\sigma_{2}^{2} \\cdot (1- \\rho^{2})}$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$k =\\sqrt{\\frac{det B}{(2\\pi)^{n}}} = \\frac{1}{2 \\pi \\cdot \\sigma_{1}\\sigma_{2} \\cdot \\sqrt{(1- \\rho^{2})}}$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$$\n",
+    "\\left(\\begin{array}{c} \n",
+    "x_{1} \\\\ x_{2}\n",
+    "\\end{array}\\right)\n",
+    "\\left(\\begin{array}{cc} \n",
+    "\\sigma_{2}^{2} & -c\\\\\n",
+    "-c & \\sigma_{1}^{2}\n",
+    "\\end{array}\\right)\n",
+    "\\left(\\begin{array}{cc} \n",
+    "x_{1} & x_{2}\\\\ \n",
+    "\\end{array}\\right) = x_{1}^{2}\\sigma_{2}^{2} - 2 x_{1} x_{2} c + x_{2}^{2}\\sigma_{1}^{2} =\n",
+    "\\sigma_{1}^{2} \\sigma_{2}^{2} \\left( \\frac{x_{1}^2}{\\sigma_{1}^{2}} + \\frac{x_{2}^2}{\\sigma_{2}^{2}} - 2 \\rho \\frac{x_{1}x_{2}}{\\sigma_{1}\\sigma_{2}} \\right)\n",
+    "$$ "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$\\Phi (x_{1}, x_{2}) = \\frac{1}{2 \\pi \\cdot \\sigma_{1}\\sigma_{2} \\cdot \\sqrt{(1- \\rho^{2})}} \\exp \\left( -\\frac{1}{2 \\cdot (1- \\rho^{2})} \\left( \\frac{x_{1}^2}{\\sigma_{1}^{2}} + \\frac{x_{2}^2}{\\sigma_{2}^{2}} - 2 \\rho \\frac{x_{1}x_{2}}{\\sigma_{1}\\sigma_{2}} \\right) \\right)$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Substitue $x_{i}$ with $x_{i}- a_{i}$ to achieve the exspected form."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 3",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.7.4"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}
+%% Cell type:markdown id: tags:
+
+### 1 Exspectation Value and Variance of a Sample
+
+%% Cell type:markdown id: tags:
+
+1)
+
+%% Cell type:markdown id: tags:
+
+See lecture slide 46.
+
+%% Cell type:markdown id: tags:
+
+2)
+
+%% Cell type:markdown id: tags:
+
+$\left< V(x) \right> = \left< \frac{1}{N} \sum_{i} (x_{i}-\mu_{x})^{2} \right> = \frac{1}{N} \left<  \sum_{i} x_{i}^{2} - 2 x_{i} \mu_{x} + \mu_{x}^{2} \right> = \frac{1}{N}  \left( \sum_{i} \left< x_{i}^{2} \right> - 2 \left< x_{i} \right> \mu_{x} \right) + \mu_{x}^{2} = \left< x^{2} \right> - 2 \mu_{x}^{2} + \mu_{x}^{2} = \left< x^{2} \right> - \left< x \right>^{2} = \sigma_{x}^{2}$
+
+%% Cell type:markdown id: tags:
+
+By fixing a parameter for the measured mean like it was done in 1) one has to apply a correction factor. If the true mean is known this is not neccessary any more.
+
+%% Cell type:markdown id: tags:
+
+### 2 Poisson Distribution
+
+%% Cell type:markdown id: tags:
+
+$P_{k} (T) = \frac{1}{k!} e^{- \lambda T} (\lambda T)^{k}$
+
+%% Cell type:markdown id: tags:
+
+(lhs) $\dot{P}_{k}(T) = \frac{1}{k!} e^{- \lambda T} (- \lambda (\lambda T)^{k} + k \lambda^{k} T^{k-1} )$
+
+%% Cell type:markdown id: tags:
+
+(rhs) $\lambda [P_{k-1}(T) - P_{k}(T)] = \lambda \left[ \frac{e^{-\lambda T}}{(k-1)!} (\lambda T)^{k-1} - \frac{e^{-\lambda T}}{k!} (\lambda T)^{k} \right] = \frac{1}{k!} e^{- \lambda T} (- \lambda (\lambda T)^{k} + k \lambda^{k} T^{k-1} )$
+
+%% Cell type:markdown id: tags:
+
+$\Rightarrow \dot{P}_{k}(T) = \lambda [P_{k-1}(T) - P_{k}(T)]$
+
+%% Cell type:markdown id: tags:
+
+Differential Equation was derived by comparing small time intervall with a Taylor expansion, see lecture slide 62.
+
+%% Cell type:markdown id: tags:
+
+### 3 Bionomial- and Poisson Distribution
+
+%% Cell type:markdown id: tags:
+
+1)
+
+%% Cell type:markdown id: tags:
+
+Bionomial Distribtion for probobility for success with N trials!
+
+%% Cell type:markdown id: tags:
+
+$B(k; N, p) = \left(\begin{array}{c}
+N \\ k
+\end{array}\right) \cdot p^{k} \cdot (1-p)^{N-k}$
+
+%% Cell type:markdown id: tags:
+
+N = 60
+
+%% Cell type:markdown id: tags:
+
+p = 0.01
+
+%% Cell type:markdown id: tags:
+
+k = 0
+
+%% Cell type:markdown id: tags:
+
+$\Rightarrow B(0; 60, 0.01) = \left(\begin{array}{c}
+60 \\ 0
+\end{array}\right) \cdot 0.01^{0} \cdot (1-0.01)^{60-0} = 1 \cdot 1 \cdot (0.99)^{60} = 54.7 \%$
+
+%% Cell type:markdown id: tags:
+
+2)
+
+%% Cell type:markdown id: tags:
+
+Poisson Distribution for event which occur with an average event rate $\lambda$.
+
+%% Cell type:markdown id: tags:
+
+$P(k, \lambda) = \frac{\lambda^{k}}{k!} e^{- \lambda}$
+
+%% Cell type:markdown id: tags:
+
+$\lambda$ = $0.01 \cdot 60 = 0.6$
+
+%% Cell type:markdown id: tags:
+
+k = 0
+
+%% Cell type:markdown id: tags:
+
+$P(0,0.6) = \frac{0.6^{0}}{0!} e^{- 0.6} = 1 \cdot e^{-0.6} = 54.9 \%$
+
+%% Cell type:markdown id: tags:
+
+### 4 Poisson Distribution, Search for free Quarks
+
+%% Cell type:markdown id: tags:
+
+1)
+
+%% Cell type:markdown id: tags:
+
+$\sum_{i=1}^{110} P(i;229) = \sum_{i=1}^{110} \frac{229^{i}}{i!} e^{-229}= 1.6 \cdot 10^{-18} = p_{se}$
+
+%% Cell type:markdown id: tags:
+
+Very low propability, very unlikely to happen!
+
+%% Cell type:markdown id: tags:
+
+2)
+
+%% Cell type:markdown id: tags:
+
+N = 55000
+
+%% Cell type:markdown id: tags:
+
+$p_{me} = N \cdot p_{se} = 8.9 \cdot 10^{-14}$
+
+%% Cell type:markdown id: tags:
+
+3)
+
+%% Cell type:markdown id: tags:
+
+$\sum_{i=1}^{28} P(i;57) = \sum_{i=1}^{28} \frac{57^{i}}{i!} e^{-57}= 6.7 \cdot 10^{-6}$
+
+%% Cell type:markdown id: tags:
+
+Much more likely to happend!
+
+%% Cell type:markdown id: tags:
+
+4)
+
+%% Cell type:markdown id: tags:
+
+Combine both Poisson Distributions to achieve the new probability:
+
+%% Cell type:markdown id: tags:
+
+$\mu = 4$
+
+%% Cell type:markdown id: tags:
+
+$\mu \lambda = 229$
+
+%% Cell type:markdown id: tags:
+
+$\sum_{i=1}^{110} \sum_{N=0}^{\infty} P(i;N\mu) \cdot P(N;\lambda) = \sum_{i=1}^{110} \sum_{N=0}^{\infty} \frac{N \mu^{i}}{i!} e^{-N \mu} \cdot \frac{\lambda^{N}}{N!} e^{-\lambda} = 4.2 \cdot 10^{-5}$
+
+%% Cell type:markdown id: tags:
+
+Even more likely to happen.
+
+%% Cell type:markdown id: tags:
+
+### 5 Why are soccer results more random than handball results?
+
+%% Cell type:markdown id: tags:
+
+1)
+
+%% Cell type:markdown id: tags:
+
+$\lambda_{1} = 1$
+
+%% Cell type:markdown id: tags:
+
+$\lambda_{2} = 2$
+
+%% Cell type:markdown id: tags:
+
+$\sum_{i=1}^{\infty} \sum_{l=0}^{k-1} P(k;\lambda_{1}) \cdot P(l;\lambda_{2})= \sum_{i=1}^{\infty} \sum_{l=0}^{k-1}  \frac{\lambda_{2}^{k}}{k!} e^{-\lambda_{1}} \cdot \frac{\lambda_{2}^{l}}{l!} e^{-\lambda_{2}} = 18.3 \%$
+
+%% Cell type:markdown id: tags:
+
+2)
+
+%% Cell type:markdown id: tags:
+
+$\lambda_{1} = 10$
+
+%% Cell type:markdown id: tags:
+
+$\lambda_{2} = 20$
+
+%% Cell type:markdown id: tags:
+
+$\sum_{i=1}^{\infty} \sum_{l=0}^{k-1} P(k;\lambda_{1}) \cdot P(l;\lambda_{2})= 2.6 \%$
+
+%% Cell type:markdown id: tags:
+
+Summing up over all possible combinations, i=1 as at least one goal is needed to win a game.
+
+%% Cell type:markdown id: tags:
+
+### 5 Python Script (thanks to Joep Geuskens)
+
+%% Cell type:code id: tags:
+
+``` python
+from scipy.stats import poisson
+import numpy as np
+```
+
+%% Cell type:code id: tags:
+
+``` python
+def func(l=1):
+    arr  = np.array([poisson.cdf(k-1, 2*l)*poisson.pmf(k,l) for k in range(1,l*10)])
+    p    = np.sum(arr)
+    diff = arr[-1]/p          # relative size of the last term compared to the total probability
+    print(f"P(B>A)={p:.4f}")
+    print(f"Diff={diff:.2g}") # should be sufficiently small
+```
+
+%% Cell type:code id: tags:
+
+``` python
+print("1)")
+func(1)
+print("\n2)")
+func(10)
+```
+
+%% Output
+
+    1)
+    P(B>A)=0.1826
+    Diff=5.6e-06
+    
+    2)
+    P(B>A)=0.0258
+    Diff=1.9e-60
+
+%% Cell type:markdown id: tags:
+
+### 6 Multidimensional Gaussian
+
+%% Cell type:markdown id: tags:
+
+$B =
+\frac{1}{\sigma_{1}^{2}\sigma_{2}^{2}- \rho^{2}\sigma_{1}^{2}\sigma_{2}^{2}}
+\left(\begin{array}{cc}
+\sigma_{2}^{2} & -c\\
+-c & \sigma_{1}^{2}
+\end{array}\right) =
+\frac{1}{\sigma_{1}^{2}\sigma_{2}^{2} \cdot (1- \rho^{2})}
+\left(\begin{array}{cc}
+\sigma_{2}^{2} & -c\\
+-c & \sigma_{1}^{2}
+\end{array}\right)$
+
+%% Cell type:markdown id: tags:
+
+$c = \rho x_{1} x_{2}$
+
+%% Cell type:markdown id: tags:
+
+$det B = \frac{1}{\sigma_{1}^{2}\sigma_{2}^{2} \cdot (1- \rho^{2})}$
+
+%% Cell type:markdown id: tags:
+
+$k =\sqrt{\frac{det B}{(2\pi)^{n}}} = \frac{1}{2 \pi \cdot \sigma_{1}\sigma_{2} \cdot \sqrt{(1- \rho^{2})}}$
+
+%% Cell type:markdown id: tags:
+
+$$
+\left(\begin{array}{c}
+x_{1} \\ x_{2}
+\end{array}\right)
+\left(\begin{array}{cc}
+\sigma_{2}^{2} & -c\\
+-c & \sigma_{1}^{2}
+\end{array}\right)
+\left(\begin{array}{cc}
+x_{1} & x_{2}\\
+\end{array}\right) = x_{1}^{2}\sigma_{2}^{2} - 2 x_{1} x_{2} c + x_{2}^{2}\sigma_{1}^{2} =
+\sigma_{1}^{2} \sigma_{2}^{2} \left( \frac{x_{1}^2}{\sigma_{1}^{2}} + \frac{x_{2}^2}{\sigma_{2}^{2}} - 2 \rho \frac{x_{1}x_{2}}{\sigma_{1}\sigma_{2}} \right)
+$$
+
+%% Cell type:markdown id: tags:
+
+$\Phi (x_{1}, x_{2}) = \frac{1}{2 \pi \cdot \sigma_{1}\sigma_{2} \cdot \sqrt{(1- \rho^{2})}} \exp \left( -\frac{1}{2 \cdot (1- \rho^{2})} \left( \frac{x_{1}^2}{\sigma_{1}^{2}} + \frac{x_{2}^2}{\sigma_{2}^{2}} - 2 \rho \frac{x_{1}x_{2}}{\sigma_{1}\sigma_{2}} \right) \right)$
+
+%% Cell type:markdown id: tags:
+
+Substitue $x_{i}$ with $x_{i}- a_{i}$ to achieve the exspected form.
+
+%% Cell type:code id: tags:
+
+``` python
+```